Question

The eartesian equations of the line are $3x+1=6y-2=1-z$. Find its equation in vector form and find direction ratios of the line.

Answer 1

The given Cartesian eq of line

$3x+1=6y-2=1-z$

$3(x+1/3)=6(y-2/6)=-(z-1).$

$\frac{x+1/3}{1/3}=\frac{y-1/3}{1/6}=\frac{z-1}{-1}$

$\frac{x-(-1/3)}{1/3}=\frac{y-1/3}{1/6}=\frac{z-1}{-1}$

multiply denominator with 6

$\frac{x+1/3}{2}=\frac{y-1/3}{1}=\frac{z-1}{-6}.$

$\therefore$ The dr's of equation

is (2, 1, -6).

The vector from of the

equation is

$\frac{x+1/3}{2}=\frac{y-1/3}{1}=\frac{z-1}{-6}=\lambda .$

$x=2\lambda -\frac{1}{3};y=\lambda +\frac{1}{3};$

$z=-6\lambda +1$

so, the vector form is

$(2\lambda -\frac{1}{3})i+(\lambda +\frac{1}{3})j+(-6\lambda +1)k.$